int_0^{frac{pi}{2}} frac{sin^2 x}{sin x + cos | vedclass

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(B) $I = \int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx \quad ... (i)$ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા:$I = \int_0^{\

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$\frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)$

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(B) $I = \int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx \quad ... (i)$ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા:$I = \int_0^{\pi/2} \frac{\cos^2 x}{\cos x + \sin x} dx \quad ... (ii)$$(i)$ અને $(ii)$ નો સરવાળો કરતા:$2I = \int_0^{\pi/2} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} dx = \int_0^{\pi/2} \frac{1}{\sin x + \cos x} dx$$2I = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \frac{1}{\sin(x + \pi/4)} dx = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \operatorname{cosec}(x + \pi/4) dx$$2I = \frac{1}{\sqrt{2}} [\log|\operatorname{cosec}(x + \pi/4) - \cot(x + \pi/4)|]_0^{\pi/2}$$2I = \frac{1}{\sqrt{2}} [\log|\operatorname{cosec}(3\pi/4) - \cot(3\pi/4)| - \log|\operatorname{cosec}(\pi/4) - \cot(\pi/4)|]$$2I = \frac{1}{\sqrt{2}} [\log(\sqrt{2} + 1) - \log(\sqrt{2} - 1)] = \frac{1}{\sqrt{2}} \log\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)$કારણ કે $\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = (\sqrt{2} + 1)^2$, તેથી $2I = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)^2 = \frac{2}{\sqrt{2}} \log(\sqrt{2} + 1)$$I = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)$

$\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx = $

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